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Are pilots required to notify the tower when they are flying "under the hood" for IFR training?
I was reading a question about flying under the hood for Instrument Flight Rules (IFR) training when this question popped to mind.  Are pilots flying under the hood required to notify the tower?  Does it change the towers approach to that plane at all (especially in VMC conditions)?

Possible duplicate of When are pilots required to have approval to conduct practice approaches?

I asked that question referenced in the 1st comment, and I will agree that this is not a duplicate, but it is certainly related.

@RonBeyer Ryan and Ralph hit it on the head.  My question is in regards to all phases of flight, not just approaches.

No, because the responsibilities aren't any different. And, with a safety pilot on board, the capabilities really shouldn't be either.

I disagree, what about aircraft that would need to hold short of the ILS hold line rather than the runway hold short? The AIM also pretty clearly states that practice approaches should be informed to the handling airport, although the AIM is just a reference.

I've NEVER seen tower hold anybody short of the ILS critical areas for a practice approach.  Even down to 800-2, you can be flying a "real" ILS without the critical areas in effect.  Where it's most important to have the ILS protected is when aircraft are flying approaches below Cat I, where any disturbance in the ILS beam can cause abrupt control inputs from the autopilot; practice approaches (being hand-flown) are far less sensitive to these perturbations.

I've been in this scenario multiple times.  We don't need to tell them that we're doing simulated instrument training, but we do need to tell them that we want practice approaches.  I'm pretty sure that if you request a practice approach the controller can figure out that you're doing training or currency.

@RalphJ, well, I suppose if somebody requested practice Cat II (or III) approach they would, but how many GA airports even has Cat II ILS.

@JanHudec   I do hear aircraft notify tower that they're using auto-land (when the weather is well above when you'd need it), particularly in ATL, which I believe IS for exactly this reason...  no incursions into the ILS critical areas during the coupled approach & landing.  But, that's for the airplane's currency; neither pilot is at all "under the hood."

@RalphJ, yes, auto-land is Cat II. And if they forget (or tower does not approve), they end up like these. And yes, that's not under the hood. But I am not sure, may be it does count for pilot currency towards auto-land too?

@RyanMortensen That does seem like it would make some amount of difference (however minor) in the process they would normally follow on a clear day.  Might be nice if some of those considerations were put into the answer (or another answer) just for the sake of completeness.

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Why does flow separation cause an increase in pressure drag?
Why does flow separation cause an increase in pressure drag?
In this question, I am relying on this answer for some information, so it might be essential to read that answer to understand what I am talking about, but I will give a quick rundown.
In that answer, it says that when flow separates the pressure of the separated area is actually higher than if it was attached. If you apply this to a wing, the separated flow area will have a higher pressure, therefore not creating as much lift (assuming the higher pressure separated flow was on top of the wing like in a normal stall).

However, lift is directly linked to drag. So if you don't make as much lift, you don't make as much drag.
So when flow separates, it causes a loss of lift. Wouldn't this also cause a 'loss' of drag?

Let's use a cylinder as an example:

Imagine in this picture the flow was attached. According to the answer I linked, the pressure behind the object will be lower when the flow is attached. The linked answer also mentions the inverse is true, so the pressure in separated flow would be higher (although still low pressure).
If that is the case, why would flow separation cause an increase in pressure drag? If the pressure is actually higher behind an object in separated flow, why wouldn't the drag actually decrease?

We can have a look at the pressure distribution of the airfoil in attached and separated flow conditions.
This is shown in this image from Fundamentals of Aerodynamics, by John D. Anderson, 6th edition, p.391

The image shows suction downwards. In the previous question, we talked about the point of separation (i.e. the front of the airfoil). Here we see that the earlier statements were confirmed, and the pressure for the attached flow case is much lower (thus more lift) than for the separated case.
Looking at the rear of the airfoil, we see that the opposite is true and that the separated case has more suction than the separated case.
So it's not categorically true that the pressure is higher/lower for separated flow compared to attached flow, it depends on the region.

In the front: less suction for separated flow
In the rear: more suction for separated flow

This gets compounded by the direction of the local flow, see this image below, from the same source:

Here we see something interesting, the arrows are pointed more vertically in the front region, and more horizontally in the rear region.
This is due to local curvature of the airfoil:

In the front: vertically oriented arrows
In the rear: horizontally oriented arrows

We can now combine both findings:

In the front: less suction for separated flow, vertically oriented arrows -> less vertical force for the separated flow: less lift
In the rear: more suction for separated flow, horizontally oriented arrows -> more horizontal force for the separated flow: more drag

Thanks for your answer. So using this same logic, why would separation on a cylinder increase the drag? The curvature over a cylinder is constant, so the rear half wouldn’t become lower pressure once flow separates like it does with an airfoil. Basically, with a cylinder the pressure would be lower on the rear side compared to separated flow, so wouldn’t drag decrease when flow separates over a cylinder?

(Essentially all separated flow on a cylinder would be the same pressure (mostly) because the cylinder is symmetrical, unlike an airfoil)

There is one theory I have: Like how the back of an airfoil turns low pressure when it stalls instead of becoming more high pressure (like the other parts of the wing), the back of a cylinder does the same thing in a flow-separated-condition. Is this correct? ("The same thing" meaning the stalled area of a cylinder is low pressure when it stalls)

(If that doesn’t make sense I can try to explain it better)

However, lift is directly linked to drag. So when flow separates, it causes a loss of lift. Wouldn't this also cause a 'loss' of drag?

Lift is not directly related to drag: it is related to induced drag. So yes, when the wing stalls, lift reduces and consequently induced drag reduces as well (at least if stall is "uniform" along the wing and therefore the spanwise lift distribution remains more or less the same - for example it remains elliptic before and after the stall).
Anyway you're forgetting the viscous drag. When the wing stalls, the boundary layer drastically grows making the viscous drag grow as well.
Normally in aviation this second effect (the increase in viscous drag) is bigger than the first one (the reduction - if any - in induced drag) and therefore stall  produces a net increase of drag. There are anyway cases where the opposite is true and a stalled wing actually generates a lower total drag.

@RobertDiGiovanni: just corrected, thanks for the comment

Thanks for the answer. I did read the link, but I having a bit of trouble understanding why in some cases flow separation will reduce drag. Could you generally explain why in some cases drag will reduce with flow separation?

@RobertDiGiovanni As said in Sophit’s answer, in some cases flow separation will reduce drag (which you suggested they add to the answer). Is there any good explanation for why in some cases it will reduce drag? Essentially I am asking you the same thing as my other comment on this answer.

@RobertDiGiovanni I think I understand that part fairly well. The part that is confusing is why in some cases induced drag will be increased instead of decreased. My main question is what differs between the case where separation increases induced drag  and when it decreases induced drag. (I do think I didn’t explain this very well in my last comment)

@Wyatt: how total drag reduces in case of a stalled F1 spoiler is nicely described at the beginning of the paper I linked: "in the case of the finite wing with low aspect ratio, the induced drag is dominant and is proportional to the downforce. Therefore, flow separation on the wing increases the profile drag while simultaneously reducing the induced drag, resulting in a decrease in the total aerodynamic drag."

@sophit I see, thanks for pointing that out. Just to clarify, wouldn’t this make it depend on aspect ratio to determine if induced drag would increase or decrease in a stall?

(I can explain what made me think this if that helps)

Why does flow separation cause an increase in pressure drag?

The pressure drag you're talking about is only a portion of the total drag, which originates from the turbulent section of the boundary layer (TBL).

Source: Nasa.
When the angle of attack increases, the transition occurs closer to the stagnation point at the leading edge, the laminar section of the boundary layer (LBL) is reduced. After the transition the boundary layer becomes turbulent. The wide wake following the airfoil greatly changes the pressure field at the rear of the airfoil and behind it. Comparing a laminar pressure field (plain curve) and a boundary layer with a turbulent portion (dotted):

Source: ibid.
This the TBL and the induced wake which increase the pressure drag. You can see the lower pressure at the trailing stagnation point (which is no more a stagnation point with a TBL).

Source: ibid.
The pressure drag is different from the friction drag present both in the LBL and the TBL.

The friction drag is due to air viscosity: Air adheres to the airfoil, the airfoil is slowed down due to the tendency of air streamlines at different velocities to shear.

The pressure drag is not this shear effect, it results from a difference of static pressure applied to the front (stagnation pressure) of the airfoil and to the rear of the airfoil. With a boundary layer entirely laminar (e.g. at 0-lift angle) these two pressures are equal and there is no net pressure drag.
But if there is a TBL portion, then the wake creates a wide volume of low pressure at the rear of the airfoil. There is now a difference between front and rear pressures resulting in a net pressure drag which slows the airfoil.

Pressure drag is due to turbulence of different sorts

In the wake of the TBL,

for a 3D wing in the tip vortices.

So when flow separates, it causes a loss of lift. Wouldn't this also cause a 'loss' of drag?

Turbulence creates a pressure drag which doesn't exist when the BL is entirely laminar. Friction drag and pressure drag sum up. Comparing an ideal flow (all laminar) with a boundary layer where the TBL exists:

Source: ibid.
While TBL decreases the amount of lift, it increases the amount of total drag (due to the additional pressure drag). The larger the angle of attack, the longer the TBL, the larger the additional pressure drag.

Here is how I have visualized this process in the past but now I am not sure it is correct, and would appreciate comments/corrections.
Using the example above of a cylindrical object with a wake of separated flow, we can imagine replacing the wake with a protrusion of solid material affixed to the downstream side of the cylinder, producing a "teardrop" cross-section instead of a circular one. The teardrop occupies the full volume of the separated region which means the flow immediately adjacent to the wall of the teardrop shape is not separated but instead follows the contour of the teardrop. Any drag effects caused by the separation dynamics are then absent because we have streamlined the shape of the body over which the flow is happening.
Now you can also see that as the flow striking the upstream side of the object splits at the stagnation point and begins moving sideways, it ballistically overshoots the actual contour of the body and fails to follow that contour- tracing out a profile that is larger in cross-section than the body itself. Inside that profile, the flow is also separated- it is as if the object presents a larger profile than its actual dimensions, and therefore more drag.
Putting a slightly sharper shape on the upstream side of the object parts the flow more gradually and yields an effective wake cross-section smaller than the case where the flow was separated.
Continuing this process of reshaping the object to eliminate flow separation all around the surface of the object finally yields an elongated object much larger in volume than the original round cross-section yet with the same or less total drag as the round cross-section.
Does this way of looking at things have any merit? Please advise.

This is a good way to interpret what happens. Proof: Take the famous golfball with and without dimples. The smooth golfball has early, laminar separation and a huge wake, the dimpled golfball a later, turbulent separation and a much smaller wake.

Appreciate the upvotes but this question deserves a better answer.
The cylinder picture in the question really tells all.

where is the center of low pressure when flow separates?

Not only how low the pressure is compared with attached flow but in what direction?
It seems clear that when flow separates the suction (center of low pressure) moves further behind the object.  In the case of the cylinder, directly behind it.
So, one can see the resultant drag vector depends not only on low pressure, but also on the direction of the pressure vector relative to the flight path of the object.
As Peter K pointed out, at a high enough Reynolds number, with flow behaving more in a invicid manner, flow will remain attached over a larger area.
Note that increase in pressure is linear, but shrinkage of the low pressure area behind the object is a square function (area) when flow remains more fully attached.

@Wyatt finding the formula for induced drag here we see L$^2$ in the numerator.  So, technically, induced drag decreases as lift decreases.

The flow past a cylinder is a good way to explain what happens with airfoils, too. So let's look what NASA has to show for different levels of viscosity:

Of particular interest is the theoretical case of frictionless flow on the upper left (case 1): The streamlines look symmetrical, and so does its pressure distribution (the dot-dashed line in the plot below, source).

This means without friction the air pressure does as much pushing from the rear as from the front. The suction around the middle is also equal, and all pressures combine to zero resulting force: Dragless flow! The same happens with airfoils in the theoretical case of inviscid flow, even as lift is produced. This is known as d'Alembert's paradox.
With a high enough Reynolds number (think millions), friction becomes less dominant and the flow becomes more similar to the inviscid case as Re increases. Typical airplane Reynolds numbers are in that region, so with attached flow there is indeed some pushing done on the rear part of the airfoil by the recompressed and still attached flow.
Flow separation, however, occurs when that recompression cannot be completed because the speed in the lower part of the boundary layer drops below zero. With incomplete pressure recovery, the rear part of the wing still experiences mild suction (the typical pressure coefficient in separated flow is -0.2) and instead of being pushed, the wing is sucked backwards a bit. This will show up as increased drag, caused by separation.
At the same time, the airfoil-boundary layer-combination will look like a wing with less angle of attack, so the suction peak at the forward upper section of the airfoil will shrink and less leading edge thrust will pull the airfoil forward. This also contributes to the drag increase with separated flow.

Thanks for your answer. So to explain this I'm going to use the case one picture. When the flow goes over the back half of the cylinder, it turns towards the centerline. When this turning is being done, low pressure is created. This low pressure is actually lower than that of separated flow, so wouldn't drag actually decrease when flow separates on a cylinder? I'm going off of this answer when writing this.

@Wyatt Pressure is proportional to the distance between streamlines. If the streamline pattern is identical, so is the pressure. And around the two stagnation points (1 forward, 1 backward), the distance is largest, and so is pressure. None of the pictures in the other answer show inviscid flow, and low speed, fully laminar flow is very different from inviscid flow.

I think my confusion is because I'm not thinking of inviscid flow. I'm thinking of a flow that (theoretically) stays attached but that isn't inviscid. If you look at the first picture in @ROIMaison's answer to this question, it shows a graph illustrating what I'm attempting to explain. The pressure with attached flow is lower, which would mean attached flow behind a cylinder (in non-inviscid flow) would have a lower pressure than separated, causing more drag to be made in non separated flow.

(if that doesn't make sense I can attempt to explain it better)

@Wyatt The lines in ROIMason's pictures are not streamlines but isobars. BIG difference! Also, the first picture shows low Reynolds number flow, like dropping a marble in honey. This is very different from what hapens in air.

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How exactly does the startup vortex form, and why does it occur when circulation around the wing changes?
So I was wondering how exactly the startup vortex forms, why it dissipates, and where it's located in relation to the trailing edge of the wing. Also, why does a startup vortex form when circulation around the wing changes? (see quote below)

"This starting vortex formation occurs not only when a wing is first set into motion, but also when the circulation around the wing is subsequently changed for any reason whatever." Millikan, Clark B., Aerodynamics of the Airplane

If someone could explain that to me, I'd appreciate it, thanks
(If someone could also explain how the tip vortices contribute to the startup vortex, I’d also appreciate it)

(What I meant by startup vortex ^)

It's written just a couple of line before in the same book. I'd also suggest you to use something more modern

Maybe this answer or this one help?

@sophit thanks, those answers did help a bit, but I'm still kind of stuck on why exactly the vortices stop being formed when at higher speeds, and why changing circulation will make startup vortices. I read the lines in the textbook you provided but I am still a bit confused.

It's just a matter of Newton's second and third law (aka momentum conservation). If the circulation around the airfoil changes (in time or space) then a proportional and opposite circulation is released in the wake.

@sophit Ah okay, so what direction and where will the startup vortex formed from changing circulation be?

"proportional and opposite": if the vortex on the airfoil is clockwise (lift upward) then the starting vortex is counterclockwise. If the (upward) lift increases, another counterclockwise vortex proportional to the increase is released in the wake. And viceversa.

@sophit ah, I see. thanks for your help!

Does this answer your question? Why is the stagnation point on the upper surface before the start of circulation?

Useful visualization

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Why are shockwaves allowed different angles at different distances from a body?
It's been (well) established that a shockwave can only be at the angle of the Mach cone. (I now understand this part). However, in this picture:

you can see that the shock is at 2 angles. I also understand why this happens. Mainly because of the 3D relieving effect, and also because the shock can never be at a greater angle than the Mach cone.

The question:
Why is the shock allowed to essentially have 2 angles in this case? Towards the rear and further away from the object, the shock is at a different angle than it is near the object. I understand why this happens, but I don't understand why a single continuous shock is allowed to have 2 different angles. Essentially, because the shock can only ever be at the same angle as the Mach cone, why is it seen at 2 different angles in this picture, considering the Mach cone is only at one angle in this picture?

Another example:
Here is a picture of a T-38C:

If you look closely, near the plane some of the shockwaves are at different angles than the others. When they get further out, they all tend to the same angle (the Mach cone). Why are the shockwaves allowed to have different angles close to the body? Why aren't they forced to the Mach cone angle, like the shocks are further out?

Chart for shock angles:

You know the answer to this.
The shock angle is determined by how much it turns the flow.  Recall the oblique shock chart I've posted multiple times -- we call it the theta-beta-Mach chart.  Theta is the turn angle, beta is the shock angle, Mach is the upstream Mach number.
Different places around the body need different amounts of turning.
The 3D relieving effect means you need less turning away from the body than near the body.
Edit:  Include oblique shock chart.

I see. I'll ask this elsewise: Imagine the Mach cone is say 30 degrees. The shock further away from the body will also be 30 degrees. Now closer to the body, the shock is at 35 degrees. (See first picture for example) In the picture, the shock is clearly allowed to exist (the one closer to the body), why is that considering it's at a higher angle than the Mach cone? Asked differently, why doesn't the Mach cone force the shock to be 30 degrees? Is it because of the different flow around the object, compared to freestream?

If the Mach angle is 30 degrees, consulting the theta-beta-Mach chart, we see that the freestream is Mach 2.0.  For a Mach 2.0 freestream and a 35 degree shock, we see that the turning angle is approximately six degrees.  So, near the body, where the shock is 35 degrees, the flow is turning six degrees.  There is no inconsistency.

Care to post the chart you mentioned into this answers to avoid the need for searching?

@Jpe61 I just edited it into my question

@RobMcDonald I see. So basically the flow turning angle is always the same, so the shock angle will never violate any rules, if that makes sense?

The flow turning angle is whatever it needs to be.  This might be caused by the immediately local geometry, other nearby flow features (fans, shocks),etc.  But together, the turning angle and upstream Mach determine the shock angle.  Period.  Full stop.  When you look at the chart, the zero-degree deflection point for each Mach number curve is the Mach angle.  All the finite deflection points are higher angles, so yes, the 'rules' are never broken.

@Jpe61 Sorry for leaving it out.  It is just the Oblique shock chart.

Sorry for nagging about it @RobMcDonald  I just think it is essential the answers work as "stanalone". The subject of aeromystique is very broad, and especially newcomers get easily lost in a maze of rabbitholes if answers become too dependent on eachother.

@RobMcDonald I see. I think it might even be more correct to say that the flow turning angle can never be greater than the Mach cone. (Correct me if I’m wrong)

@Wyatt  That is not a useful comparison.  Draw a ramp and a shock.  Of course the shock is outside the ramp.  Let's say your upstream Mach is 2.0.  From the chart, the maximum turning angle is about 23 degrees.  What happens if you hit a 35 degree ramp with a Mach 2.0 flow?  (I know the answer, do you?)

@RobMcDonald That’s a good question. My guess is it would turn into a 90 degree normal shock?

That is correct.  It is forced to a normal shock -- and the flow downstream of a normal shock is subsonic (the flow downstream of an oblique shock is still supersonic, just slower).  That subsonic pocket of flow can then deal with whatever the obstacle is.


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What determines the angle at which a shock will form?
What determines the angle that a shock will form? If you look at a lambda shock, you’ll see it isn’t straight up, but with an angle (both the front and rear part of the lambda shock).
The second comment on this answer helped me, but what is meant by “turning”?
Basically I was curious why some shocks are curved, others are straight up, (relative to the wings surface) and some are straight but angled, (also relative to the surface of the wing) if that makes sense. Thanks.

The angle of an oblique shock is determined by the Mach number before the shock and how much the flow turns.
Usually, this is illustrated by a ramp.  A 10deg ramp means the shock turns 10 degrees.
This is an oblique shock chart, it illustrates all oblique shocks for a perfect gas.

There are two solutions -- a strong shock and weak shock solution.  The weak shock solution will have supersonic flow downstream of the shock.  The strong shock solution will have subsonic flow downstream of the shock.
Nature generally prefers the weak shock solution unless it is forced into the strong shock solution.
So, a Mach 1.6 flow that must turn 10deg will have a shock wave angle of about 51 degrees.
This all makes sense when the upstream flow is uniform, and the downstream flow must turn the same amount.
However, imagine a situation where the flow had to turn 10deg, but the upstream flow wasn't uniform in Mach number -- lets assume it varied from 1.6 to 1.8.
Each individual streamline has its own oblique shock situation -- an intermediate streamline at Mach 1.7 that must turn 10 degrees will have a local shock angle of 47 degrees.  The streamline at 1.75 that must turn 10 degrees will have a shock angle of 45.5 degrees.
Put all this together and you get a shock that is curved.
A curved shock results from a flow that has either 1) non-uniform Mach before the shock, and/or 2) the need to turn a non-uniform amount.
Edit: Stuff below about flow turning added.
Here is a typical diagram used to explain oblique shocks.  You will find this diagram and the shock chart above in any compressible flow textbook.
Region 1 is upstream of the shock.  Region 2 is downstream.
Notice how the velocity vectors in region 1 are parallel to the wall at the bottom.  They must be -- otherwise, the conservation of mass would be violated.
The wall bends at point A.  It turns by angle $\theta$.
Notice how the velocity vectors in region 2 are parallel to the wall at the bottom.  Once again, they must be -- otherwise, conservation of mass would be violated.
So, the sharp bend in the wall forces the flow to turn.  This in turn forces the shock to exist.  The strength of the shock -- and the shock angle are determined by the upstream Mach number and the turning angle.
The thing is, this $\theta$-$\beta$-$M$ (pronounced theta-beta-mach) relationship is true whether the wall is there or not.  I.e. if you have a flow of a given Mach and a shock of angle $\beta$, then it will locally turn by angle $\theta$.  Or, if you have a flow of a given Mach that turns by angle $\theta$ for whatever reason, the local shock angle will be $\beta$.

Edit: Adding lambda shock image to support comments.
The first shock of a lambda shock is the one that causes the separation.

Ah I see, thanks for that graph; I couldn't find a good result myself. So I'm not sure if this requires a separate question, but what is meant by flow turning? Where does it turn? Sorry if I'm not picking up on something obvious.

I have extended my answer to address your question, but this really is trivial and would have been answered if you had read any resource on oblique shocks.  Turning the flow is the entire point of an oblique shock.  Once again, I'm going to ask you to put in a bit more effort on your end rather than just asking here.

I see. So because the shock is there, it forces the flow to turn, resulting in the shock turning? (I usually just search to an extent (typically on google), but if no good results show up I just assume that there is no good results to my question. Now that I say it, it's a dumb assumption, so I'll definitely try searching more before asking here)

Compressible flow is all about the boundary conditions.  Something forces the flow to turn, so a shock happens.  If it turns the other way, an expansion happens.  Causality is from turning to shock.  However, if you see a shock, you can know that turning has occurred (and you can know how much if you know the Upstream Mach and the shock angle relative to the flow).

Ah okay, that makes sense. Last little thing I was curious about : Would you consider the shock on the top of the wing a type of oblique shock? (The one that forms on the upper surface, about halfway down depending on the speed)

Nevermind, I researched it a bit more and found what I was looking for. A normal shock is 90 degrees to a surface and an oblique shock is what we were just talking about. That would make a lambda shock a type of oblique shock, correct? (I will delete unnecessary comments afterwards)

lambda shocks are made of multiple oblique shocks.  A normal shock is a special case of a oblique shock.  A normal shock is always the strong shock solution, but turns zero degrees.

On top of the middle of a wing, the surface is smooth (curved, but no sharp corner), so it won't be forced to turn -- the velocity immediately before and after the shock are parallel -- so that is a normal shock.  Once the shocks reach the trailing edge and they must turn the flow to make the corner, they will transition into an oblique shock.

I see, thanks. Last question : if normal shocks form on the top of the wing, then why is the front shock on a lambda shock considered an oblique shock? I know that normal shocks are a type oblique shock, but what differs to make an oblique or normal shock, on the top of the wing?

If the shock causes separation such that the first streamline outside the boundary layer needs to sharply turn, then there will be turning and the shock will clearly be an oblique shock.  However, if there is no such turning, then the shock can be considered a normal shock.  A normal shock is just a strong oblique shock with zero turning angle.

Ah okay, I get it. So I’m assuming there’s a second shock somewhere to make the separation happen in order for the oblique shock to form.

No.  The first shock in a lambda shock would be the one to cause the separation.


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Why does the angle of a shock change when the flow leaves the influence area of an object?
In this (great) answer, it says:

"far enough" from the Concorde's nose, the path followed by the airflow is not affected by its presence and it's just a straight line. Close to the nose we need a shock wave at 45° to make the airflow turn the 15° of the nose. Anyway, a bit further, the flow turns a bit less than 15° (let's say 10°) and therefore a shock wave of 40° is enough to achieve this inclination (point b). Still a bit further, the flow turns again a bit less (let's say 5°) and therefore a shock wave of 35° is this time enough (point c). And still a bit further, we reach the "far enough" distance where the airflow is just going straight: we reached the point c where θ
is zero (i.e. the airflow goes straight) and the shock wave is 30°.

So the oblique shock is made because the flow needs to turn rapidly. It starts from the object turning the flow, then the shock made from that turns the flow and makes another bit of shockwave, then that process repeats. (See here for a more comprehensive explanation)
After the flow leaves the influence area of an object, why is this shock angle able to change? What makes it so the shock can achieve the same turning angle with a different shock angle?
Here is a photo to illustrate what happens after the flow leaves the influence area of an object:
On the right side, you can see the shock bends once it leaves the influence area. That shock still needs to achieve the same amount of flow turning, but it's at a different angle. What allows it to have a different shock angle but still have the same amount of flow turning?

Basically, the shock needs to turn the flow the same angle no matter what. What allows the shock to turn the flow the same amount even though the shock itself is at a different angle?
The best reason I have thought of is because the flow directions/patterns are different around the object, so the shock has to be at a different angle to achieve the same amount of flow turning (relative to the airflow).

I just re-read the answer I provided at the beginning of this question, and found a good thing to illustrate my question.

approaching the pointy nose of the Concorde the airflow must turn 15° and this is achieved via an oblique shock wave at 45°. The further we are from the nose, the smaller the turn of the airflow is and therefore the smaller the inclination of the shock wave is.

"The further we are from the nose, the smaller the turn of the airflow is and therefore the smaller the inclination of the shock wave is"
Why is the turning smaller further from the nose?

The 'influence area' of an object in supersonic flow extends downstream at the Mach angle -- forever.  Likewise, the area that can influence a point in supersonic flow extends upstream at the Mach angle -- forever.
I think your mental model of the 'influence area' is somehow limited to the vicinity of the body -- that is incorrect.

It seems we have 2 phenomena at work here, air being moved by the energy of the moving object and an energy pulse being transmitted by the air: the sound wave.
At the leading edge interface the (supersonic) object both moves the airmass and initiates the sound wave.
Behind the object (on the right), the low pressure area moves air inwards, resulting in a bending of the wave inwards.
Once the wave leaves the "influence area" it propagates away at the same angle.
The relationship of mass flow and energy waves can be seen analagously in tidal bores, which can make a wave moving into an estuary stationary because the outflow of the river matches the inward velocity of the (transmitted) wave.

Downvoter please step forward, identify yourself, and read this reference.  What do you think?

It seems they downvoted my question also (I think). +1 from me though.


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Do seaplane have designated landing areas?
Do seaplanes have to land in particular places or are they allowed to land pretty much anywhere there is room to land safely?

See: If I have a float plane, can I land on any body of water?

Taken from seaplanes.org:

While most pilots assume the FAA has jurisdiction over landing areas, including water-based landing areas, the truth is much more complex. Jurisdiction rests with the person or organization that "owns" the waterway. This may be a Federal or state agency, a local government, a private corporation, or an individual. Determining who controls a waterway is the first step in determining whether it is legal to land on that body of water.
A second complication enters in the picture with overriding jurisdictions, most commonly state-imposed seaplane base licensing requirements. In several states, notably Ohio, New Jersey and Indiana, seaplanes may not land unless the proposed landing area is certified as a seaplane base, regardless of whether the waterway owner provides permission or not. To determine whether this is an issue in your area, call your state aeronautics office (often a division of the state department of transportation), check the Water Landing Directory, or call SPA Headquarters (863/701-7979).

As to WHERE on the body of water you're allowed to land, I have no clue.  Interesting question, though.

Huh, yeah, that would make it especially hard to figure out landing on a river like, say, the Mississippi (which is what I was thinking of), the particular part of the river I want to land on is between Illinois and Iowa...  I guess I just have to carefully pick a side and call that state...for starters.

From what i understand yes.
They land on waterways

TITLE XXXIX
  AERONAUTICS
  CHAPTER 422
  NEW HAMPSHIRE AERONAUTICS ACT
  Prohibitions and Penalties
  Section 422:27
      422:27 Seaplanes in Operation on Public Waters. –
I. All seaplanes shall be considered boats while in operation on the waters of the state and shall be subject to the marine rules of navigation, except that they shall be exempt from all laws and rules concerning the operation of boats for the purpose of landing and taking off from such public waters.
II. The operation of seaplanes shall be subject to any restrictions placed upon the use of public waters by rules adopted by the department of safety or the department of environmental services.

Source. 2002, 6:1, eff. July 1, 2002.
If you are using your plane as a boat you could effect another view. Such as fishing and bring a citation from Fish and Game or Marine patrol. The NH RSA. is clear any state water body, May be used for take off and landing without being subject to additional boating regulations. Additional restrictions are hard to see from the air , but apparently can exist. By local town ordnance and state management areas.
On another note as of 2018 it seem NH fish and Game has made changes to the state wide boating access FIS 1600 . Restricting the use of their controlled public launch and parking No person shall land or launch and aircraft or drone from.

In addition to the above post NH considers water bodies of 25 acres or more to be the property of The State . the largest lake in NH is also under US Coast Guard jurisdiction and may have additional requirements. The grey area  is still speed limit restrictions .

This doesn't really answer the question. It specifies that they're like boats when they're on water, but not which waters they're allowed to use, and doesn't explain what the implications are of being "considered boats"

In the State of NH there are designated seaplane area on some lakes . The NH RSA  allows for any State water body to be used for a landing and take off. The designated seaports are landing areas for seaplanes with additional boating restriction benefits such as fuel available. They also eliminate speed restrictions that may otherwise be in question under navigation requirements.

Hi Norman, welcome to Aviation SE. Please edit your answer and add the information given in your last comment, ideally with some link to original sources.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Why does the twin otter have wing fences?
Why does the twin otter have wing fences? Planes like the Cessna 172 don't have them, so why does the twin otter have them? (I know that it is used for STOL purposes, but how would wing fences help in that way?)
My guess is that they're for helping with stalls, but how would wing fences help with a stall? (This picture is of an RC twin otter, couldn't find a good real-life one)

Why there's shouldn't be any spanwise flow on a unswept wing? You have even asked about it in another question of yours!?!

@sophit the spanwise flow I was talking about there is a pretty minor effect, read the second comment on this answer to see what I mean. Since that effect is pretty minor, I didn’t see a purpose in wing fences. I’m still going to edit this question, because you’re right about me contradicting myself.

Here is a real-life photo, although a poor one, plus an unauthoritative answer.  Briefly: spanwise flow does happen on the Otter; the fences avoid stall where the ailerons are, esp. when ailerons droop along with flaps.  That preserves roll authority on approach and flare.

@Wyatt spanwise flow is minor at full speed and minimal angle of attack, in "normal" airflow. When a part of a wing ceases generating underpressure (stall) while the other part continues, it becomes major enough. Also, during normal operation overwing spanwise flow is towards center, and when the center stalls, it reverses.


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Why will air keep accelerating in Mach 1+ flow, when it would decelerate in subsonic flow?
So I was clarifying a few things in a comments section of an answer, and I had quite a few questions so I thought I'd ask my main one in a separate post.
In the second comment on this answer, it says that air will continue to accelerate in supersonic flow, where it would decelerate in subsonic flow. I understand the part about disturbances reaching further out, but what would make air continue to accelerate in supersonic flow, when it wouldn't in subsonic flow? Also, this affects the pressure distribution of the wing, but why doesn't it affect the whole wings lift distribution evenly? (The answer was super helpful, but I didn't understand this part)
Thanks!

For supersonic flow to accelerate, it must have a negative pressure gradient  in flow direction. The acceleration is "powered" by a loss of pressure and will end if pressure downstream is higher. In that case, air will abruptly decelerate in a shock.
In transsonic flow you will have accelerating air in a small supersonic pocket, followed by a shock. Ahead of the shock pressure is low so there is a pressure gradient which drives the acceleration. Typical examples are supersonic regions on the upper side of wings or convergent-divergent nozzles where the speed of sound is reached at the end of the convergent part ("throat") and the divergent part accelerates the now supersonic flow further. Once the flow crosses into the freestream at the end of the nozzle, it is recompressed in a sequence of shocks.
Now you might ask why the flow will have a negative pressure gradient when outside pressure is higher. In supersonic flow there is no mechanism which allows the higher outside pressure to travel upwards into the divergent nozzle - only when the flow exits the nozzle will that pressure travel into the core of the supersonic stream with the speed of sound, leading to the diamond pattern of shocks. Ahead of those shocks it is only the nozzle geometry which controls whether and how much the flow expands and accelerates.

ah okay. So basically the shockwave on top of the wing acts kind of like a barrier of sorts, shifting the lift distribution forward?

I’m not 100% familiar with aerodynamics at Mach speeds, so that’s just what I interpreted it as.

@Wyatt I have to admit that I do not understand your comments. Keep in mind that small pressure changes cannot travel upstream in supersonic flow.

ah okay I'll try to explain it better.  My thinking is that the shockwave where the pressure recovers over the wing will be like a wall, to where the air gets rapidly decelerated, changing the pressure distribution.

@Wyatt Yes, of course. The shock causes an increase of pressure, but if it is heavy enough, also a flow separation downstream. The supersonic area ahead of the shock causes a drop in pressure compared to the fully subsonic pressure distribution. See here or here for more.

oh okay thanks. Why does the lift distribution at the tips change differently than the rest of the wing with transsonic speed? In your other answer, it mentions that but I didn’t see why. Sorry for asking so many questions and this comment being kind of off topic, I don’t want to ask tons of separate questions on here, if that makes sense.

@Wyatt The question was about wingtips, so the answer was for wingtips. I suggest you read a bit more about fundamentals instead of asking imprecise questions.

Yeah you're probably right. I probably will write a separate question for the part about wingtips at transonic speed. (Of course along with reading more basic fundamental things)


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Does the bending air on a wing create one or multiple vortices?
As explained in the first paragraph of this answer, air will bend inwards on the top of the wing, and outwards on the bottom of the wing. This creates an inward flow component on the top of the wing and a outwards component on the bottom. I'm pretty sure this would create vortices/vortex once the wing has passed. It sounds like what I'm saying might be hard to understand, so maybe try to visualize the flows I was talking about.
So my main question is this : Would this bending of air on the surfaces make one big vortex or multiple smaller ones once the wing has passed? The picture on this answer might help convey what I mean by air bending on the upper and lower surfaces.
Also, how would tip vortices change this, since they contribute to these spanwise flows?
(If that doesn't make sense, I will happily clarify in the comments and edit this question if needed)

You have already got a very good answer to the present question in another similar question of yours: do you actually read the answers that you get?

@sophit yeah I recalled reading that answer, but it doesn’t really state anything about how many vortices are created along the wings trailing edge.

How would you count a distributed vortex?

One might imagine one giant vortex rolling off the back of a straight wing like a carpet, and more vorticies off a swept wing.  A highly swept wing may have fewer, larger vorticies, but more parallel rather than perpendicular to the fuselage.

I personally like the picture used on Wikipedia. It explains in a simple way where the vortices form and how the coalesce into the tip vortices.

@sophit oh okay, so the vortices I was talking about combine into the tip vortices? Neat.

Yes, small vortices coalesce.  Just like small individual thermals below roughly 500 ft AGL coalesce into ones big enough for sailplanes to circle in.


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Why do flat wings create a vortex/vortices?
I've been researching this for the past ~45 minutes, and I still can't find a solid answer. Why do flat wings create a vortex/vortices? In this answer, it says that the detached flow will create a vortex (at high AoA) instead of re-attaching, but I still don't really understand why and how the vortex is created.
I get why high sweep wings, like the Concorde's will make a vortex, but I don't see why a flat wing would make one. Also, do flat wings make multiple vortices across span for the same reason, or just one? (It's just confusing because why doesn't an airfoil-shaped wing make the same vortex?
That also makes me wonder, what direction does the vortex face? Does it face the same direction as a tip vortex, or perpendicular to that direction?
Thanks a lot!

That vortex is called Von Karman vortex and is typically formed behind stalled body.

@sophit I see. What might make a stalled body create a vortex?

Detached flow is turbulent.  You might as well ask, how could turbulent flow NOT have eddies and vortices.

@CamilleGoudeseune well yeah true, but I was curious why the main vortex is created on flat wings, like what causes it.

By "flat wing" do you mean "flat plate" like the answer you cite, or do you mean "unswept" like how you say that it's the opposite of the Concorde?

@CamilleGoudeseune I meant a wing like a Cessna 150, but without the airfoil curvature.

@Wyatt straight wings at their most efficient angle of attack do not utilize vortex lift.  Chaotic vorticies in turbulent flow are useless, a sustained large vortex, as seen on delta wings  does generate lift, but even that is not nearly as good as attached flow on a cambered wing.  Vortex generators are only helpful at high angles of attack to keep airflow attached.

At a sufficiently high angle of attack, a rectangular thin-plate "airfoil" suffers flow separation at its upper surface,
just like any conventional airfoil would.
(A thin plate is not at all special in this regard.  It just has poor L/D and
poor structural stiffness compared to conventional airfoils.
Its post-stall behavior hardly differs from that of other airfoils, cambered or not.)
This separated aka detached flow is turbulent.  Turbulence has vortices.  That's all there is to it.
Some physicists even define turbulence as
"when an ordered fluid flow breaks into small vortices."
Questions about how or why turbulence causes vortices, or whether turbulence just is vortices,
would be better for the physics SE, or maybe even the philosophy SE.
Based on observation, the largest vortices will be spanwise, but small chordwise vortices also appear,
and everything in between.  This is turbulence, after all.  It's messy.

how a vortex is created

One key ingredient is sustained wind shear.  We can see this on wingtips as high pressure is drawn up and over the end of the wing.
This flow is in a different direction than the low pressure area and the free stream, creating a rolling motion driven by the energy of the motion of the aircraft.
Notice that not only the degree of sweep affects vortex formation, but also the 2 dimension shape of the airfoil.
The delta wing (a paper airplane favorite) is essentially two wingtips put together, making a strong lifting vortex.
A straight wing works differently, using lift created by attached airflow for much more efficient lift.  Here we don't want vorticies, which is why straight airfoils are thicker and rounded.
"Vortex generators" can increase the critical stalling AoA of a straight wing (anything is better than separated airflow), but at a cost of higher drag.
While deltas are versatile at many speeds and can be made thin like flat plates, straight airfoils are far better lifters.
A straight wing can produce a vortex perpendicular to the wing tip vortex at its trailing edge, but this is only when at a very high drag, impractical angle of attack.  The ones vortex generators make are parallel to the wing tip vorticies.

I think you might be confused about what I'm asking. I was asking why a wing without the curvature from an airfoil has a vortex in the front of it. Imagine a piece of ply-board being used as a wing. I think this answer would definitely help convey what I mean. (read middle - end of the first paragraph)

@Wyatt  The picture in your referred answer is showing (in blue) an area of lower pressure, not (necessarily) a vortex.  But notice how tiny it is compared to an airfoil.  But remember, on a paper airplane scale and speed (Reynolds number), most of the lift comes from higher pressure on the bottom "surfing" the air.  This is why toy gliders can have thin wings (which actually lower drag).  One of the great discoveries of around 105 years ago was that once you go faster, attached flow provides fantasticly higher lift/drag ratios.  This lead to thicker wings on monoplanes.

@Wyatt please read the end of the second paragraph of your referenced answer.  It seems to link the vortex behavior of the tip of the straight wing and delta wings.

hmm, I could see where a vortex might be created, but the one I'm thinking of would turn in a direction perpendicular to they way tip vortices turn.

@Wyatt yes, that would be at the trailing edge.  The most desirable, lowest drag way to make lift is to have clean (laminar) flow until the wing is past.  This is not unlike the end of a power boat separating from its wake.  The increase in speed from less drag is noticable.

okay, thanks. To clarify, the vortex illustration in this picture at the front of the wing is what you're thinking of?

@Wyatt the plate in the picture will producing lift from the bottom.  Those little curlies on top most likely represent flow separation.  Useful lifting vorticies (from vortex generators or with deltas) will run parallel to the fuselage, but again, are way less efficient.

@Wyatt there is a type of airfoil called the Kline-Fogleman that attempts to make use of vorticies that are perpendicular to the tip vorticies by "stepping" the upper surface of the wing.  Jury is out on their performance.


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Do 2 vortices next to each other have any attracting force?
Do 2 vortices next to each other have any attracting force?
Say you have 2 tip vortex sized vortices next to each other. Will they have any sort of attracting force toward each other? If they do, why is that?
(For simplicity's sake, say a spacing between the 2 vortices of whatever makes sense for explaining)
Thanks and happy new years!

I’m voting to close this question because this has been asked and answered before. The author is advised to use the search function and to actually read the multiple answers.

@PeterKämpf would you mind telling me which question has answered this one? I mean I always search for similar questions before I ask them

Was this question closed because off topic or duplicate? If off topic then I'd propose to reopen it because it's a legitimate aerodynamics question. If duplicate then I don't see any link to the duplicated one...

@sophit I’m not sure of any other questions that answer this specific one. I am clueless as to why this got marked as off topic.

@Wyatt: Frog's comment to the answer of How do 2 vortices (or more) combine when they are near each other?

@Peter oh I remember reading that comment. It helped a lot, but it didn't seem like a full answer, so I thought I'd ask a separate one. If the answer to my question is as simple as that comment explains it as, then I get closing this question.

@Wyatt A vortex is not a physical entity - it is just a flow pattern which we separate from the rest of the flow and give a name to. Please stop thinking of it as an object separate from the rest of the air.

@PeterKämpf yeah, that helps to think of it as just a flow pattern. Thanks for your help!


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What causes the lambda shape in a shockwave?
What causes the λ (or lambda shape) in a shockwave? I heard it had something to do with the interaction between the shock and the boundary layer. If so, how do they interact and why does it make the signature λ symbol?
If someone could explain it in an easier way to understand, I’d appreciate it, thanks.

I heard it had something to do with the interaction between the shock and the boundary layer

I will edit this question to add the link to where I saw that. Can’t seem to find it right now.

Did you try googling 'lambda shock'?  I suggest you do an image search, there are a lot of explanations out there.
Here is an interferometry image.  Not as pretty as a shadowgraph or Schliren, but still very informative.  If you google around, you will find many more -- but often of even more complex phenemona.

Region 1 is upstream of the shock (C1).
Shock C1 causes an adverse pressure gradient that causes a flow separation bubble.  Flow separates at S, but reattaches at R.
Inside the bubble (the streamline connecting S to R), there is a recirculation zone.  Flow outside the bubble keeps going.
A second shock (C2) forms on the downstream side of the bubble.
These shocks meet at point 1, and a combined shock (C3) extends to the freestream.
The surface $\Sigma$ is a slip line between zones 3 and 4.  Velocities will be parallel across that line.  Pressures will be equal across the slip line.  Temperature, density, entropy, etc. will be different across the slip line.
You should check out An Album of Fluid Motion.  Image 246 for example.
Also, here is a great video (from a computation) of flow in a shock tube.
Initially, flow is everywhere at rest in a section of tube.  The top of the tube is far above us (we are zoomed in on the bottom wall).  The right-side of the tube is closed.  To our left, there is a membrane separating a region with 100x the pressure.  That membrane is popped and the shock starts propagating down the tube (to the right).
When the shock passes a location, it induces a velocity to the right.
Eventually, the shock hits the wall at the right of the tube, it reflects back to the left.
Now, instead of entering a region of stationary air, the shock is encountering a flow.  Since the bottom wall of the tube is there, the flow has built a boundary layer along the bottom wall.
The shock causes the boundary layer to separate, which forces some turning -- forming an oblique shock on the bottom wall that forms a lambda structure.

Ah I see, thanks. So not sure if I’m just not picking up something, but why do the shocks have the angles that they do? The first one has a backward angle, and the other one has a forward angle.

The angle of a shock depends on the upstream Mach number and the amount of turning.  You can look up (google image search) an oblique shock table and it will show this relationship.  This chart can be derived from basic compressible fluid equations.

Oh I see. I thought it might have something to do with how the shock is formed. The last thing I am curious about : I thought for a shock to form the flow has to be supersonic (not an oblique shock) and the flow behind the first shock is subsonic? So how would the second one form?

The flow in front of a shock must be supersonic.  After a shock, the flow can be subsonic or supersonic.  All of the flows in the diagram above meet this requirement.  Flow is from left to right.

Oh okay, thanks. So the flow behind a shock isn't always subsonic, got it. I'm also assuming that the shocks formed don't always take on the form of a lambda shock?

Lambda shocks are a special case where multiple shocks combine together to form this kind of shape.  It forms from shock - boundary layer interaction.

Ah I see. I think that will be a good thing for me to research more. (BL and shockwave interaction). Thanks for your help!

Good luck.  While I encourage you to learn about it for your own enrichment, it is a very small corner of aerodynamics that most professional aerospace engineers (myself included) don't know much about.  If you are fascinated by it, then by all means, dive in.  Otherwise, prioritize and stay a little more shallow and explore the breadth of aerodynamics a bit more.

Yeah it seems like a hard thing to understand. I think I'll just research and question it until I have a basic understanding of what happens. (Enough to understand other topics in aerodynamics)


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Why don't shockwaves expand rearwards when they expand out?
Why don't shockwaves expand rearward when they expand out? In this for some reason really big picture, you can see the shocks expanding out past the actual plane. Why don't they also expand rearward? And for that matter, forward?
If it did do this, I'm guessing it'd make something that looked like a triangle.

You're going to have to clarify what you mean by 'expand rearward' and 'expand out'. Your question doesn't make any sense to me. Note that this image is a 2D perspective on a 3D phenomena. The actual main shock at the nose of the aircraft is really a cone. The imaging technique used to generate this picture mostly captures a slice -- but with some averaging of the 3d behavior into that slice.

oh okay, will do. Basically, because shockwaves are high pressure, I assumed that it'd want to expand, like any high pressure gas does.

Also, expanding out meaning going further from the plane, like seen in the photo. Expanding rearward meaning the shock expanding into the red triangle area.

Shocks are discontinuities.

A shock isn't a region of high pressure, it's a discontinuity in pressure (see my answer).  It's a contour line on a map, not a collection of molecules whose volume could increase.

A shock is an abrupt (thin) change in properties.
As you get far from a body, the need for the shock may dissipate and the shock may cease to exist.
However, when there is a shock, it is thin -- by definition.

It might help to understand that a sound wave (better example: light wave) is an energy form that does not simply dissipate quickly by equalizing pressure).  It takes much longer, unfortunately resulting in sonic booms.

A shock doesn't expand (spread out), simply by definition.  The shock is where pressure and density change discontinuously.  If that gets smeared to cover the red triangle superimposed on the photo, then it becomes continuous, so it's no longer a shock.
A related kind of spreading is a Prandtl–Meyer expansion fan, but that may be fancier than what this question is asking.

The short answer is because the thing that is making the shock wave is moving faster than the shock wave is able to expand.
That's why a shock wave forms, rather than sound rippling out in all directions -  It's caused by all the ripples piling up.

Ah okay, that makes sense. So basically the air can't expand backward because it's moving forward so quickly?

Well, the sound is expanding backwards relative to the plane, so you will hear it after it passes, but there is no rearward expanding shockwave.

Oh okay. So last question because I don’t want to keep asking questions, but if sound can expand backwards, why can’t the shockwaves?

Because it's being pushed forward continuously?

"Expand" is being used in different ways here.  Are there better verbs we could use?

@CamilleGoudeseune, I'm open to suggestions, but that is the verb used in the question...

@Wyatt, I guess another way to think of it is that if it "expanded" it wouldn't be a shock wave any more.  The shock wave is compressed sound waves - "expand" the shock wave and you just have sound...

@MichaelHall Yeah by definition that would be true. I also asked this question, and it really helped me figure this out!


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Do the oblique shocks at the trailing edge cause as much resistance as the ones on the surface of a wing?
Shockwaves formed on a wing will move backward until they reach the trailing edge, when they turn into oblique shocks and stop. When they're on the surface of the wing, it makes sense why they would make resistance, because they're acting like walls in a way. When the shocks reach the trailing edge, they turn into oblique shocks because the flows on the top and bottom of the wing are coming at each other at an angle and they have to turn, making a shock to do that; explained very well by this answer if I didn't make sense.
For me it doesn't make sense why the oblique shocks would make resistance for the plane at all. In the same way the wake of a plane doesn't cause it more resistance, why would the shocks make resistance? They still act like a wall (they being the oblique shocks), but the oblique shocks are kind of like a byproduct in a way, so why would they make resistance?
Basically, I don't see anything that would be 'pulling' the shocks with the plane.
Here's the oblique trailing edge shocks I was talking about. Another way to think of it : The shocks as you can see are at the very tip of the airfoil, so why would it make much resistance at all?

(If any of that doesn't make sense, comment and I will try to fix it. I found it hard to explain what this question is asking. Thanks)
edit : Could someone explain why this question is getting downvoted?

The oblique shocks don't cause resistance.  However, they do cause a jump in pressure.  That jump in pressure (on the body) is what causes the change in force.
The oblique shocks at the TE do not change the force on the body.
Here is a supersonic wedge airfoil.

The shock causes the pressure in region 2 to be higher than freestream (region 1).  Then, the expansion fan makes for lower pressure in region 3.  The final shocks make for high pressure downstream of the airfoil (not depicted).
The high pressure on the front compared to the low pressure on the back cause  what we call wave drag.
All aerodynamic forces are either pressure or shear.
In an intermediate transonic case, with a shock on the top surface, like this one...

The shock causes the boundary layer to separate -- and that region of separation on the back of the airfoil is what really causes the drag.
From this, we see that transonic drag rise has a totally different physical origin than supersonic wave drag.

That’s very interesting. So the majority of the drag from the shocks isn’t always from the shock acting like a wall?

The shock acting like a wall causes the pressure increase.  The pressure increase either causes the drag -- or causes flow separation, which causes the drag.  All aerodynamic forces are pressures or shear.

Oh okay that makes sense, thanks for your answer. Also do you know why this question got downvoted? I'd like to improve it, but I need to know why it needs improvement.


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Can the F-35 (and F-22) do the Kvochur Bell maneuver?
Can the F-35 (and F-22) do the Kvochur Bell maneuver? Surprisingly I didn’t find much about this on google. (Might’ve searched for the wrong thing though)

F-22 has thrust vectoring, so it most likely is able to do whatever. F-35 does not, so its ability to do fancy circus tricks is somewhat limited.

There's YouTube video of a Su-27 performing the maneuver. The nozzles aren't moving much if at all. It doesn't look like TVC is required. I haven't found a good description of the maneuver, but it looks like most maneuvering happens pre-stall, the back half of the maneuver is just a tail slide.

Come to think of it, the Su-27 doesn't even have TVC.

@JörgWMittag yeah those are some good points. I would love to see an F-35 do that maneuver! I think a lot of people would be surprised if it could.

@Wyatt, here's a good video of a Raptor performing a J-turn which is similar at:
https://www.reddit.com/r/aviation/comments/tqgc2c/f22_pilot_demonstrates_a_minimum_radius_turn/?rdt=63687

The F22 fighter jet can do a tailslide, which the bell maneuver is a variation of. Here's a video (can only be viewed on YouTube):

The embedded video doesn't appear to be viewable without switching to YouTube?

I'm surprised by this, as answers on other StackExchange sites do have embedded videos, so I guess this is not a Youtube-wide policy. Can users choose to allow embedded video's when they upload?

Yeah, whether a video is playable in embed mode depends on the uploader's settings, and also sometimes country of origin, I believe.


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Why don’t closed wings eliminate tip vortices?
It’s clear that tip vortices never really can be prevented, but why doesn’t a closed wing do that? There is no ‘tip’ on a closed wing so where does the air go to make a vortex?
On a normal plane the air goes from the bottom surface to the top and curves around the wingtip, but what does it do in the case of a closed wing?

Closed wings are a bad idea anyway because the second wing as seen in the picture will probably create more drag than the tip vortices themselves.

I’ve also heard that a closed wing will have problems with boundary layer separation at the tips, so maybe that has something to do with it?

Related: Do box-wings suffer from induced drag the same way as normal wings?

@PeterKämpf Ah, interesting. It seems the box wing is basically the same thing as a closed wing as far as tip vortices go.

Also, if you see this @RobMcDonald, do you know why your answer was deleted? I was about to comment on it.

Find the fuel consumption and climb rate of that thing and compare it to a Piper Cub.

FYI, Rob deleted his own answer because it morphed into a discussion... (I guess you need a certain number of reputation points to see deleted answers)

@MichaelHall Ah I see. Thanks for informing me because I can't see the answer myself like you assumed.

@Wyatt Yes, I deleted it myself.

So, rivers of words have been written around this question but we don't have an answer, cool...

@sophit yeah there was an answer in the reply that got deleted, but since it got deleted you can’t see it anymore.

@Wyatt: yes I know, I can see that deleted answer. Still I don't understand why it simply doesn't get "streamlined" and undeleted since it is technically correct


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Why doesn’t the air expanding behind a shockwave reverse the effects of wave drag?
Wave drag is (from my understanding) the difference in pressure between the front and rear of an object due to shockwaves, making an opposing force. This needs higher pressure in front of said object and lower pressure behind.
Now shocks themselves aren’t high pressure, but high pressure occurs right after the shock.
(Again from my understanding), higher pressure air will expand and cause lower pressure. Same thing that happens with an airfoil.

So if wave drag is from there being higher pressure on the front of an object, and higher pressure air will expand making lower pressure air, why wouldn’t the two pressure differentials just cancel each other out?
To make things simpler just imagine you have a square plate angled at 45 degrees going into the free stream.

So to recap, the air will contact the 45 degree plate, make a shockwave which will have high pressure after it, then that higher pressure air would expand and make low pressure. Wouldn’t this high pressure then low pressure cancel the net force out? Am I missing or misunderstanding something?

My guess is that this effect does happen on say an airfoil, but by the time the high pressure air expands and turns low pressure, it's already passed the wing or where it won't change wave drag. Not sure though.
Also, since shockwaves aren't always right close to the surface, the expansion of the high pressure might still make low pressure, but it will not be contacting the object. See the second picture in the answer here to know what is meant by "the shocks aren't always close to the surface".

You're forgetting the expansion fan on the upper part of the plate

@sophit Oh okay, so how might that change the wave drag amount?

@sophit Sorry if I’m not understanding something obvious, but isn’t the expansion fan because of the low pressure? The expansion fan doesn’t do anything on its own, right?

This is what happens at supersonic speed on a flat plate

I see. So I was basically forgetting that the back side of said object is also lower pressure, so even if the front of the object had high then low pressure (like I mentioned above) it wouldn’t cancel anything out? Hope that makes sense haha

Why do you think there is high pressure after the shock?  It might help to think of it as the airfoil moving through air (and forget wind tunnel modeling for now).  The airfoil has a boundary layer which is also moving.  It might be accelerated at a curvature to Mach 1, and there is your shock wave.  The shockwave has a peak and trough just like an ocean wave.  The peak has higher pressure.  But remember, a wave is not moving air, it's energy transferred through air.  The area of compression is very thin and constantly moving.  The energy from moving the airfoil (thrust) creates the wave


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Why are the outside of the rings of an afterburner exhaust brighter that the rest?
In this answer’s comments, it is explained (with a link for more information) why afterburners have rings in the exhaust.
To summarize, the rumble of the engine + afterburner creates a resonance that causes density changes throughout the exhaust.
However I can’t seem to understand why the outer part of these discs are brighter.

If you zoom in, it’s clear that there are these ring structures in the exhaust. My question is why is the outer part of these discs brighter (AKA higher density)? Shouldn’t the pulses be equal throughout each disc? (It looks more equal when you look at the ones closer to the nozzle)
The document linked in that answer goes over the different frequencies that cause this effect, but I couldn’t seem to find anything on why the outer part of the disc is brighter.

The best theory I have so far is that the outer flow of the exhaust has more unburnt fuel, but I wouldn’t know why that would be the case.

It also seems that the outer part is (close to) the same brightness as the inner part on the rings closer to the nozzle.

The air pressure causes the exhaust to collapse in until it gets too compressed and air pressure isn't strong enough anymore to squeeze it. It's still hot and energetic, so it rebounds back out until it's equal to the air pressure. At that point, the air pressure starts squeezing back in again, causing the exhaust to collapse again, and so on. The bright points in the exhaust cone are the spots where the exhaust is squeezed in due to atmospheric pressure.

@AadirajAnil It sounds like you’re describing Mach diamonds. The last few comments of the first linked answer should clear things up.

“I have to guess, but the jet looks like it has almost atmospheric pressure and the rings are the consequence of density changes resulting from oscillations ("afterburner rumble") in the afterburner pipe. So they are not actually shock diamonds.” - One of the comments on the linked answers talking about this same effect.

I'm guessing it's the same phenomenon responsible for smoke rings.

@MichaelHall Good point. It seems that this subject isn’t documented very much from what I’ve researched.

When you say "outer part of the ring" do you mean the existence of the rings themselves?  i.e. Why not a flat glowing disc with each pulse, why do they manifest as a ring?

@MichaelHall I made an edit changing just that. (I think it sounds right, but if not let me know!)

OK, I don't know enough about mach diamonds to merge the info into a good and complete answer, but for more information to back my hunch above, they would seem to be a vortex ring:  https://en.wikipedia.org/wiki/Vortex_ring

The brightness in the diamonds/rings is not due to external combustion in the flow. The light comes out of the afterburner and is refracted by the external flow. The angle of refraction varies with gas pressure (influencing the refraction index). This reveals where the shock disks are located. The flow itself produces no light.

@mins That sounds very plausible. I had not thought of that before, and that would explain why only the outer part of the discs are super bright.

I'm not sure this explanation is complete thou, it's possible carbon particles and/or water droplets are also glowing due to pressure and temperature as explained here (simple black-body radiation without combustion). Also be aware of IR being captured as white by many sensors. This Q/A on space.se about cryogenic plumes (not colored) seems to confirm both effects.


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When an oblique shock forms, what happens to the normal shock that helped it form?
What happens to the normal shock that helped the oblique shock form, in the first shock of a lambda shock? Oblique shock waves form because :
"An oblique shock wave is a shock wave that, unlike a normal shock, is inclined with respect to the incident upstream flow direction. It will occur when a supersonic flow encounters a corner that effectively turns the flow into itself and compresses." - Wikipedia.
(Over this whole question, I'm referring to the first shock in a lambda shock, just to make an example)
For the oblique shock to form, there has to be a normal shock first, I think. The normal shock makes a separation bubble which makes the flow turn, making an oblique shock. My question is, what happens to that normal shock when the oblique shock forms? Does it disappear? If so, why? This answer and its comments has some really good info, but (for me) it didn't explain this part.
Also, if the flow behind a normal shock is always subsonic, how would the oblique shock form? I thought flow had to be supersonic to form an oblique shock?

Does this answer answer your question?

@ROIMaison Thanks for the link, but it doesn’t really talk about how the oblique shock I was asking about forms, unless I missed something.

The way I read it, initially there is a normal shock, which disrupts the flow and causes separation. This separation bubble presents an obstacle to the flow, which means it needs to generate an oblique shock to get around it.

@ROIMaison That’s actually how I thought of it too, but my question was what happens to that initial normal shock?

@ROIMaison (what happens to it when the oblique shock is formed of course)

I found a schematic of the situation with a separation bubble in the paper Capsule aerodynamics and shock-wave boundary layer interaction (SBLI) in supersonic and transonic flow :

This shows that the normal shock is still there, but the surface to which it's normal is not the airfoil, but the displaced surface of (airfoil + separation bubble + boundary layer).
This makes sense, to the flow, the separation bubble + boundary layer is also an obstacle that it has to navigate.

I see, thanks. So if the air behind a normal shock is subsonic, how do the oblique shocks form? I thought flow had to be supersonic for an oblique shock to form.

Is it because by the time the normal shock creates enough separation for the oblique shock to form, the flow is still supersonic?

In case of a normal shock:
Supersonic-shock-subsonic (look at the value of M2 in the tables wiki Normal shock In case of an oblique shock: it depends on angles and speeds, see how this image has two solutions, one for a strong shock M2<1 and one for a weak shock M2 >1.
Oblique shock graph
The oblique shock solution was also explained to you here

Ah I see. I think I understand it more. So for an oblique shock to form the flow has to be supersonic, so in some fashion the normal shock is still able to form the oblique shock with supersonic flow. (Probably doesn't make sense how I explained it)

It's the other way around. Supersonic flow leads to shocks. The flow is supersonic and needs a method to get back to subsonic. The shocks are the way to do that. If the flow is going straight, you get a normal shock, if the flow needs to turn, you get an oblique shock. Then there are two options, either there is a strong shock that directly makes the flow subsonic, or there is a weak shock, which leaves the flow supersonic. This then is followed by another shock to return the flow to subsonic. That is how you get the double shock pattern, the lambda.


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Is my understanding of density/speed correct?
So I asked this question, and I wanted to know if my guess below was correct. I got an answer in the comments, but the reason for asking this question is because I was still a bit confused if my thinking was correct, if that makes sense. (the below is copy/pasted from the linked question) Also, if this question is considered a duplicate or goes against any guidelines, I'll happily delete it. (I did read this page, on what questions to ask and what to avoid.)

My guess of what happens with density at higher and lower speeds is the following (my guess is that it's a non-linear relationship) : The air density doesn't change much until you reach Mach ~0.3, where Bernoulli's equation isn't valid anymore, so the density can't change but so much, or Bernoulli's equation would be false, even below Mach 0.3.
When you go faster, there are less air molecules to 'bounce' off of the surrounding body and each other (because at high speed, the surrounding air is less dense), so changing speed doesn't do as much as it would at a lower speed, if that makes sense. (I might be wrong about some of those statements.)

I think if I explain my guess of why density doesn't change much up until Mach 0.3, it might clarify things. So if you have normal-density air (like still air at room temp) and you start to speed it up, a few things happen.
The air molecules bounce off of each other less, because they all have a forward speed component now, and they 'bounce' less on the surrounding container/body, both creating a lower density air. Doing this at lower speeds, there are more molecules, and because there are so many, speeding up doesn't prevent the 'bouncing' as much as it does at higher speeds. (Again, I might be wrong about some of those statements.)

Going back to basics may help here.  If air molecules "bounce off each other less", the air is cooler.  Pressure is a function of Density and Temperature.  I would not conflate this with Trans-sonic "Gedanken experiments" for now.

@RobertDiGiovanni ah okay I see. So I’m kind of confused about how the Gedankin experiment answer ‘works’ because you made a good point about pressure vs. temperature.

I know it can be confusing because, even in a wind tunnel, "wind flow" is not entirely accurately portrayed (but good enough to get some solid research done).  In reality it's the air reacting with the wing in motion.  This is why they say air isn't compressed (at lower Mach), it just gets moved out of the way.  The Gedanken experiment seemed to deal with pressure differences behind the shockwave at transonic speeds.  This took a lot of years for many people to figure out.

@RobertDiGiovanni hmm okay. I came across this answer, and if you look at the bottom of it, you'll see a part about density. Could you explain what it means, because it says "A speed increase is always coupled to a decrease in density." which is confusing  because at this point I'm not sure if density increases or decreases with speed. Thanks

density where?  I believe they are talking about phenomena behind the shockwave.  But is density vs velocity a function of air itself?  Or is it (as pressure, Bernoulli) that a moving air (or water) stream draws air away from a surface, creating lower pressure, ergo density?  Air simply moves from high pressure to low.  To have a significant change in density (along the path of the airstream, within a few meters) would require one heck of a pressure differential.

Though one might see something like that at ... the exhaust of a jet engine.  More food for thought.

@RobertDiGiovanni hmm good points. One question, why would the pressure be lower behind a shockwave?

probably many theories, but to me a trans or super sonic shockwave is like a moving wall of compressed air, literally leaving a "wake" of lower pressure behind it within its cone.  The lower density behind it (and the higher pressure in front) may be one reason why longitudinal stability "diverges" at higher Mach.  Good luck with your studies on this very complex subject.

@RobertDiGiovanni Yeah I kind of visualized it in my mind, which helped. I can definitely see a shockwave acting like a 'wall', which behind it is low pressure. Thanks!

The density doesn't go down. If anything it goes up. Here's some intuition for the process:
The aircraft is constantly trying to mush the air particles together, which would increase the air density.
When the aircraft is very subsonic, the particles around the ones being mushed together get out of the way. This keeps the density from increasing much at all. But as it gets faster and faster the particles have less chance to get out of the way, so they bunch up instead, which results in increased density.

... to add this answer, the ability of the particles to "get out of the way" is a function of their average molecular motion (think how fast they are moving around in the fluid from just bumping into one another, the motion due to their heat or temperature), and this remains the overriding determinant up to and until the aircraft true airspeed exceeds the average thermal velocity, (transonic, or about 0.92Mach), where shock waves begin to form. Shock waves are the result of the particles inability to "get out of the way" of particles bumping into them.

Okay, thanks. Also what if there wasn’t a body or an object in the flow? It would still decrease density as it sped up, right?

@Wyatt No, it doesn't decrease whether a body is in the way or not. Density is a function of how many air molecules are in a given volume. Speeding air up has no effect on its density.

@Chris oh okay. Could you read the 3rd paragraph of this answer to see why I thought density decreases with speed? You might see why I’m a little confused because of what the linked answer says. Thanks.

Also if you see the first answer in the link I provided in the question, you might be able to see where I'm coming from more (plus the link in my previous comment)


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Why does density decrease most above Mach 0.3, and not as much below Mach 0.3?
Why does density decrease more above Mach 0.3, and not as much below Mach 0.3? From what I've heard, it isn't a linear relationship but why is that? Asked differently, what makes it a non linear relationship?
Thanks!
Link to chat room of this question.
Also, the last paragraph of this question has my guess of why the density doesn't change much below Mach ~0.3. It might be useful to see where I'm coming from.

Some reading here.

@RobertDiGiovanni Interesting concepts in there. Thanks

Q.1 asks what it does.  Q.2 mentions the definition of nonlinearity, and then asks why.  Q.3 asks the opposite of Q.1.  The third paragraph asks yet another question.  Help!  Use fewer words, not more.

@CamilleGoudeseune hmm could you quote which question is asking something opposite? I'd like to improve what's wrong with this question. I do see something that does kind ask the opposite of what I said, so I'll change it now. Any better?

Your edit just now improved the question, thanks!  But if you can edit it down even farther so it asks just a single question, that would help a ton.  Find the one best way to phrase the question, and leave it at that.  Repeating with variations is useful for poetry and theology, but in the hard sciences it's only confusing.

@CamilleGoudeseune Oh okay, so do you suggest that I ask a separate question for the 2nd and 3rd paragraphs?

Ask one question per stack-exchange-question.  How you do that is up to you!

@CamilleGoudeseune Oh okay, gotcha. I will definitely do that, thanks!

Let us continue this discussion in chat.

@CamilleGoudeseune, okay, sent you a message there.

Take the [tour] and look into how to ask a good question.

When you combine the equations for isentropic flow
you can plot
"density compared to density at zero speed, M=0", i.e. $\rho / \rho_0$, as a function of M.
Anderson's Fundamentals of Aerodynamics
presents that as
$\rho_0 / \rho = (1+ \frac {\gamma -1}{2} M^2)^{\frac{1} {\gamma-1}}$ with $\gamma = 1.4$ for air  (page 567, eq. 8.43).
Inverting,
$\rho / \rho_0 = (1+ \frac {\gamma -1}{2} M^2)^{\frac{1} {1-\gamma}}$.
I've copied a plot
of this below, but you can calculate it yourself if you like.  The red curve shows that M = 0.3 isn't magical.  (At M=0.3, the density ratio is 0.956.)  It's just a convenient, practically useful threshold for everyday calculations.
Why isn't the red curve a straight line?  Because these equations have proven to be a useful, accurate, predictive model of fluid flow in many fields, and these equations are nonlinear.  You could replace them with linear ones if you liked, but the calculations you'd make with them would be less accurate.

Anderson's own plot, p. 573 fig. 8.6, agrees:


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Why is a thicker boundary layer more turbulent?
Why is a thicker boundary layer more turbulent? I’m mainly referring to the effect of the varying speed over the boundary layer, and why it makes it more turbulent. I know it has something to do with the Reynolds number, but what happens physically to make a turbulent boundary layer?

A short answer to the question is: Viscosity.
Boundary layer is absolutely a viscous phenomenon, it arises due to "friction" between fluid (air) and surface of the wing, this might be thought of as the moving air loses some of its energy while contacting the surface, so some of the velocity is gradually lost..
When you say "varying speed over the boundary layer", if you mean the gradual vertical decrease in velocity, this is because the rate of change of velocity with respect to height over the surface is linearly proportional to the shear stress that appears due to friction, according to this law:
$\tau = \mu \frac{\partial u}{\partial y}$
Where $\tau$, the shear stress, is typically a positive value (in direction of positive x-axis), so the velocity $u$ increases gradually as height above the surface increase, until the velocity equals the freestream velocity $u=u_{\infty}$.
This is also explained in the image:

Oh okay. But why does the pressure gradient make turbulence?

Sorry for the late comment, I believe the main answer is still: viscosity, and hence, friction,
I think it's intuitive to think of it this way: if an airfoil is extremely smooth, streamlines of air would move on a path exactly on the surface of the airfoil, taking exactly its shape, this is called a laminar flow.

Friction on the other side, causes changes in the air flow, streamlines exchange velocity with each other, both velocity and direction of streamlines changes in a chaotic manner. That's why there's no theory that actually models turbulence as far as I know until today, because it's chaos.
This is how I think of it, it might not be very accurate physical explanation, but it's just my intuition and I hope it helps...

Thanks for your intuition! There's also another question I asked here that got some good answers.

Thank you, by the way when I said: streamlines exchange velocity, this is not accurate, I meant layers of air above the airfoil change velocity, but anyway you got the point

Ah I see, thanks for pointing that out!

A boundary layer starts at thickness zero and grows from there. Since the transition from laminar to turbulent flow can be anywhere between the start of the boundary layer and a local Reynolds number of several million, turbulent layers tend to occur when the boundary layer has already reached some thickness. Turbulent boundary layers will not go back to a laminar state, so the thickest part is almost always turbulent (except for small Reynolds numbers which on the external surfaces can only be found on small model airplanes - there, a boundary layer can stay laminar until its separation).
Next, turbulence lets a boundary layer grow more quickly than laminar flow. Actually, the question should be reversed: A turbulent boundary layer is thicker.

Oh okay, I see, thanks. One small question : why do boundary layers turn turbulent as they travel over a wing/surface?

I edited my question to capture more of what I really was asking, so that might help to see where I am at.

@Wyatt You should really ask a new question instead of re-writing one that has already an existing answer. Now the answer is off-topic.

I just realized that, thanks for pointing that out. I'll edit it back and ask a separate question for that.


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Why does the density decrease equal the speed gain at Mach 1?
In this answer, it says : "At small Mach numbers, changes in speed cause negligible changes in density, but as Mach approaches unity, both are of similar magnitude. With Ma>>1 , changes in density will become dominant. A speed increase is always coupled to a decrease in density."
Why does the density decrease equal the speed gain at Mach 1? Is it just a coincidence or a result of the Mach number? If it's not a coincidence, what makes this happen? I welcome a non-mathematical explanation, instead more of explaining what happens to the air molecules.
(it says at Mach 1 density changes will become dominant, but the 3rd paragraph of this answer might clarify what I mean)

@CamilleGoudeseune, I think this is a bit too strict of a definition. I'm pretty sure there are ample questions here that are about pure aerodynamics as opposed to applied aerodynamics (and of which it thus could also be said that they belong to Physics.SE, as you could switch the fluid)

But, coincidentally enough, this question was asked before on Physics.SE - Can Mach number be greater than 1 if mass flow rate is decreasing?.
The formula in the accepted answer explains the different behaviours at M<1 and M>1, and how therefore mass flow should be maximum at M=1

@ROIMaison, fair enough.  I'm unvoting (if that is a word!) to close.


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Might a large commercial jet dump fuel if an engine were on fire?
In my mind this seems like a bad idea for a lot of reasons, like the dumped fuel lighting on fire itself, or perhaps leaving fuel vapor in the wing tanks (which may be even worse than regular liquid fuel) or even just because if you do put the fire out you would now have a new emergency.
But, I've been wrong before.  Perhaps it would actually help put out the fire and being forced to glide would be an improvement over an engine fire, right?
So I thought I'd simply ask the experts.  Might a large commercial jet (like a B747, A380, etc) dump it's fuel if an engine caught on fire in order to starve the fire of fuel?

PS- For the sake of this thought experiment, pretend the fire had robbed the engine of the ability to cease fuel flow to it...  Unless that really makes no sense, in which case please feel free to call me on it and that can be the answer.

Your PS makes no sense; fuel shutoffs, like practically everything on a plane, have redundant backups, so even if the fire has damaged the fuel pump to the point it's leaking fuel into the engine to sustain the fire, you can cut off the fuel further back, even if that means cutting the fuel to the other engine on that wing in a four-engine config. If there's no way left to shut off the fuel, you likely have bigger problems as it would mean the damage causing the engine fire has also severely damaged the wings and fuselage.

@KeithS My PS makes sense!  But mostly because my PS specifically says that if my idea doesn't make sense please correct me ;), so, overall, as a post script, it makes sense.  Even though my idea on fuel cut offs is quite clearly wrong -_-;;

It is unlikely that dumped fuel will ignite, unless it is sprayed in the direction of the actual fire.  And if it ignites, it is unlikely to increase hazard to the plane, since the plane will be moving hundreds of knots faster than the dumped fuel.

RELATED:  https://aviation.stackexchange.com/questions/71905/why-didnt-philippine-airlines-113-dump-fuel-prior-to-forced-landing

An airliner experiencing an engine fire might dump fuel, but not to starve the engine. That is much more easily accomplished by simply shutting down the fuel pumps to that engine, which would be a first-line firefighting technique as it's the most economical and the most likely to allow for a restart (less maintenance is required on an engine that's simply shut down versus one that's been doused with chemical suppressors, and those same chemical suppressors might extinguish the fire but would also inhibit relighting it).
There are several redundant fuel shutoff systems in most airliners, from the fuel pumps themselves to actual shutoff valves in the fuel lines in case the pumps themselves are the failure point. Any failure that would completely prevent the crew shutting off the fuel supply to a damaged engine would likely create much more immediate problems, like severe structural damage to the wing or fuselage of the aircraft. All of these shutoffs in addition to active fire suppression systems such as chemical extinguishers are under the flight crew's manual control; an engine on fire is usually still producing at least some thrust, which may be more important than extinguishing the fire in certain critical phases of flight.
The plane is more likely to dump fuel if it's close to its departure airport when the emergency occurs. In such a scenario, the aircraft is likely above its maximum landing weight (which is less than the max takeoff weight as the forces on the landing gear and the runway are less on takeoff than for landing and the plane is expected to burn most of its fuel weight by the time it gets where it's going), and therefore the fuel dump is simply to reduce the aircraft's weight for an emergency landing, so the landing gear doesn't collapse on touchdown. Additionally, if anything else goes wrong during the landing, dumping fuel would leave the plane with as little fuel as possible to minimize the spread or intensity of a fire on the ground, which can impede or prevent escape from the plane. However, fuel dumps from commercial aircraft are rare, used in situations where the necessity of dumping fuel to land safely and spare the passengers outweighs the financial costs of losing the fuel and the environmental costs of cleaning up the spill.

Good answer and I up-voted it. However, I slightly disagree with: " However, fuel dumps from commercial aircraft are rare and used only in dire emergencies when the aircraft has to get back on the ground ASAP."
If it's a dire emergency, something like a bomb scenario immediately after takeoff, we were trained to make an overweight landing.
Dumping fuel takes time, which is fine if it's not a dire (severe?) emergency, at least in a 747. The one time I did it was for an engine fire and shutdown on departure. We used 20 minutes or so necessary to fly to a dump area and then dump 150000 lbs..

My understanding is that fuel dumped from an airliner evaporates, or at least stays suspended in the atmosphere until it does; it doesn't typically reach the ground.  In that case there isn't really any "cleaning up" that can be done, and in particular no monetary cleanup costs incurred (there is of course the intangible environmental cost of the pollution).  Is that not correct?

@NateEldredge I seem to remember being told in ground school that Jet-A evaporates within 2,000 vertical feet.

... Where it would contribute a significant fraction of the smog-producing VOCs in the air for miles around. No liquid may hit the ground (and even if it did it would vaporize fairly quickly) but it's an uncontrolled release of atmospheric pollutants which the airline usually has to answer for if the dump was performed within a national jurisdiction (it's possible a jet could fly out to international waters, dump, then circle around).

@KeithS - "it's possible a jet could fly out to international waters, dump, then circle around" ...and then you get SwissAir 111